Search found 48 matches

by joydip
Fri Dec 16, 2016 9:36 pm
Forum: National Math Olympiad (BdMO)
Topic: BdMO National 2016 Secondary 3: Weird angle condition
Replies: 3
Views: 1922

Re: BdMO National 2016 Secondary 3: Weird angle condition

A synthetic solution : Let $ Q$ be a point such that $\triangle PQB $ is equilateral ,$C$ and $Q$ are on the same side of $PB$.Then $Q$ is the center of $\odot BPC$. So, $AQ$ is the perpendicular bisector of $BC$ . As,$\angle APB=150^{\circ}$, so $AP \perp QB \Rightarrow AP$ bisect $\angle QAB$ .So ...
by joydip
Sat Dec 10, 2016 11:49 am
Forum: Number Theory
Topic: USAJMO 2016
Replies: 1
Views: 1571

Re: USAJMO 2016

$2^{20} \parallel 5^{2^{18}}-1$ .Then $5^{2^{18}+20} \equiv 5^{20} ( mod 10^{20})$. Now,$\lfloor {20\log 5} \rfloor+1=14$. So $ 5^{20}$ has 14 digits. So $5^{2^{18}+20}$ has exactly six consecutive zeros. Now ${2^{18}+20}< 10^6$ can be easily proved.
by joydip
Tue Nov 29, 2016 11:38 pm
Forum: Number Theory
Topic: Infinitely many primes divide $1!+2!+\cdots +n!$
Replies: 1
Views: 1197

Re: Infinitely many primes divide $1!+2!+\cdots +n!$

Let's prove a more general result: For a prime $P$ ,define $C_p = \{x : P^x \mid S_n , $ for some $n \in \mathbb N \}$, then $C_P$ is a bounded. Proof : Assume , $C_P$ is not bounded for a prime $P$. Then - Claim 1 : $P^x \mid S_{px}$ ,for any $x \in \mathbb N $ Proof: As $C_P$ is unbounded ,there e...
by joydip
Tue Nov 22, 2016 4:10 pm
Forum: Geometry
Topic: APMO 2013 P5
Replies: 3
Views: 1813

Re: APMO 2013 P5

Let, $G=BD \cap AC,F=DC \cap AQ$ .Then $,(P,G;A,C)= -1 $.So , R,F,G are collinear.Applying pascals theorem on hexagon $AEBDCC$ we get $EB, CC$(tangent at $C),FG$ are concurrent $\Rightarrow B,E,R$ are collinear.
by joydip
Tue Nov 22, 2016 12:54 am
Forum: Geometry
Topic: A Beauty from Evan Chen
Replies: 1
Views: 1231

Re: A Beauty from Evan Chen

Let $AG$ and $MO$ meet $BC$ at $K,S$ respectively .$KM$ meets $PN$ at $R$. As $M$ is the center of $\gamma$,so $MO$ is the perpendicular bisector of $AG$ . $ \angle PAM =\angle PGM =90^o \Rightarrow AP\| BC \Rightarrow \angle APM =\angle OSN.$ $ON=\frac {1}{2} AH =AM $ and $\angle ONS =90^o =\angle ...
by joydip
Mon Nov 21, 2016 11:44 pm
Forum: Geometry
Topic: Two triangles and three collinear points
Replies: 1
Views: 1889

Re: Two triangles and three collinear points

$AD$,$BE$ and $CF$ concur on the orthocenter $H$ of $\triangle DEF$. Let $D_0,E_0,F_0 $ be the projection of $D,E,F$ on $EF,FD $ and $DE$ respectively .Let $\alpha$ be the circumcircle of $\triangle DEF$. Lemma : The perpendiculars to $BC, CA, AB$ through $ R,S,T $ respectively intersect at $H$ Proo...
by joydip
Sat Oct 15, 2016 7:06 pm
Forum: Algebra
Topic: Polynomials without real solutions
Replies: 1
Views: 2026

Re: Polynomials without real solutions

My solution: Let $P$ and $Q$ be the graph of $P(X),Q(X)$. $P,Q$ doesn't intersect (as $P(X)$ can't be equal to $Q(x)$).Assume $P(P(X))=Q(Q(X))$.Consider the points $A(P(X),P(P(X))),B(P(X),Q(P(X))),C(Q(X),P(Q(X))),D(Q(X),Q(Q(X)))$. $AD \parallel BC \parallel XX^{'}$ and $AB \parallel YY^{'} \paralle...
by joydip
Sun Sep 11, 2016 11:44 pm
Forum: Geometry
Topic: Cyclic intersections of $AH$ and $BO$ gives similar $\Delta$
Replies: 2
Views: 1359

Re: Cyclic intersections of $AH$ and $BO$ gives similar $\De

$\angle XAC =\angle OAB=\angle ABX$ .so $\odot AXB$ touches $AC$ at $A$ .Similarly $\odot BYC$, $\odot AZC$ touches $AB$ & $BC$ at $B$ & $C$ respectively .By Miquel's theorem $\odot AXB$ , $\odot BYC$ & $\odot AZC$ pass though a common point, say $P$ .(Using directed angles) $ \angle XPY = \angle XP...